   # The pH of a solution of Ba(OH) 2 is 10.66 at 25 ℃. What is the hydroxide ion concentration in the solution? If the solution volume is 125 mL, what mass of Ba(OH) 2 must have been dissolved? ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 16, Problem 14PS
Textbook Problem
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## The pH of a solution of Ba(OH)2 is 10.66 at 25 ℃. What is the hydroxide ion concentration in the solution? If the solution volume is 125 mL, what mass of Ba(OH)2 must have been dissolved?

Interpretation Introduction

Interpretation:

hydroxide ion concentration in the solution has to be calculated and mass of Ba(OH)2 must be dissolved in 125mL solution also to be calculated.

Concept introduction:

The pH is a measure of hydrogen ion concentration. The expression for pH is,

pH=log[H+]                                                                                                       (1)

The sum of pH and pOH is equal to 14 at 25 °C.

pH+pOH=14                                                                                                         (2)

Molarity (M) is the concentration of solution expressed as the number of moles of solute per litre of solution.

M=n(mol)V(L)                                                                                                            (3)

Number of moles (n) is calculated as,

n=m(g)Mw(gmol1)                                                                                                   (4)

Here, (m) is the mass of the substance and (Mw) is the molar mass of the substance.

### Explanation of Solution

Barium hydroxide is a strong base, it completely dissociates into ions in an aquoeous solution as follows:

Ba(OH)2Ba+2+2OH

Given:

The pH of solution is 10.66.

The volume (V) of solution is 0.125L.

Substitute the value of pH in equation (1) to calculate pOH.

14=10.66+pOHpOH=3.34

Now substitute the value of pOH in equation (2).

3.34=log[OH][OH]=4.57×104M

Since, barium hydroxide after dissociation gives two moles of hydroxide ions. Therefore the concentration of barium hydroxide is half of the concentration of hydroxide ion in the solution.

[Ba(OH)2]=4

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